Research Methods and Statistics for Psychology: cram pack

Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.

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1Formulas and methods

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

Linear regression & correlation

Sums of squares
Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
Slope
b = Sxy/Sxx
Intercept
a = ȳ − b·x̄
Correlation
r = Sxy/√(SxxSyy)
Residual
e = y − ŷ

Chi-square tests

Test statistic
χ² = Σ(O − E)²/E
Goodness of fit
E = n·p, df = k − 1 − (estimated parameters)
Independence
E = row total × column total / grand total, df = (r − 1)(c − 1)
Critical values (α = 0.05)
df 1: 3.841, 2: 5.991, 3: 7.815, 4: 9.488
Decision
reject H0 if χ² > χ²crit

2Practice problems

1.Research methods in psychologyMedium

A researcher compares children's aggression before and after television arrives in a remote town. The researcher did not control the arrival of television. This is a:

  1. ACorrelational survey of television viewing
  2. BField experiment
  3. CLaboratory experiment
  4. DNatural experiment
2.Research methods in psychologyHard

In multiple regression predicting exam scores from study hours and sleep, R² = 0.36. What does this mean?

  1. AStudy hours cause 36% of exam success
  2. BThe predictors together explain 36% of the variance in exam scores
  3. C36% of students passed the exam
  4. DThe correlation between the predictors is 0.36
3.Descriptive statisticsMedium
202530354045

Find the sample standard deviation s of the data: 29, 47, 43, 39, 20.

n = 5
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 50 and standard deviation 10. Find the z-score z of a score of 65.

μ = 50σ = 10x = 65
z =
5.ProbabilityMedium
n = 7k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 7? The order does not matter.

n = 7k = 3
N =
6.ProbabilityHard
€p(x)−20.7320.25180.02

A game costs €2 to play. You win €4 with probability 0.25 and €20 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€4: p = 0.25€20: p = 0.02
E(G) = €
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.7486

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7486. Find z to two decimals.

A = 0.7486
z =
8.Normal distributionHard
μ = 250x = ?97.5%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 250σ = 5P = 97.5%
x = g
9.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 30 phones, mean 107 h, known standard deviation 8 h. Model 2: 60 phones, mean 96.5 h, known standard deviation 20 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 107 hσ₁ = 8 hn₂ = 60x̄₂ = 96.5 hσ₂ = 20 h90%
upper limit = h
10.Confidence intervalsHard
σ = 28, n = ?x̄ − Ex̄ + Ex̄E = 199%

The population standard deviation is 28. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 1?

σ = 28E = 199%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 98.2 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 98.2s = 12n = 36
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 56 of 80 patients on treatment A recover, and 70 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 56n₁ = 80x₂ = 70n₂ = 120
z =
13.Linear regression & correlationMedium
xy136349561771977xy5406080

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
14.Linear regression & correlationHard
xy242345447552653xy2464550x = 5.5

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (5.5, 48) is made. Find its residual e = y − ŷ.

n = 5(5.5, 48)
e =
15.Chi-square testsMedium
lowmediumhighTotalurban43112276rural27281671Total703938147

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “rural, low”.

row total = 71column total = 70N = 147
E =
16.Chi-square testsHard
dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

A chi-square goodness-of-fit test compares observed counts in 3 categories with fully specified expected proportions (no parameters are estimated from the data). Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Research methods in psychology

Answer: D · Natural experiment

Why this is the answer

In a natural experiment the IV varies because of events outside the researcher's control. Ecological validity is high, but random allocation is not possible.

Answer

D · Natural experiment
2.Research methods in psychology

Answer: B · The predictors together explain 36% of the variance in exam scores

Why this is the answer

R² shows how much of the variability in the outcome is accounted for by all predictors together. It does not establish causation.

Answer

B · The predictors together explain 36% of the variance in exam scores
3.Descriptive statistics

Answer: s = 10.99

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Mean

x̄ = 178/5 = 35.6

Squared deviations

Σ(x − x̄)² = (−6.6)² + 11.4² + 7.4² + 3.4² + (−15.6)² = 483.2

Standard deviation

s = √(483.2/4) = 10.99

Answer

s = 10.99
4.Descriptive statistics

Answer: z = 1.5

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Standardise

z = (65 − 50)/10 = 1.5

Answer

z = 1.5
5.Probability

Answer: N = 35

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

Combinations

C(7, 3) = (7·6·5)/(3·2·1) = 35

Answer

N = 35
6.Probability

Answer: E(G) = −0.6 €

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Net outcomes

G: 2 (p = 0.25), 18 (p = 0.02), −2 (p = 0.73)

Expected value

E(G) = 2·0.25 + 18·0.02 − 2·0.73 = −0.6 €

Answer

E(G) = −0.6 €

A negative value means you lose money on average.

7.Normal distribution

Answer: z = −0.67

From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.

Area to the left of z

Φ(z) = 1 − 0.7486 = 0.2514

Read the z-table backwards

Find 0.2514 inside the table and read off z.

z = −0.67

Answer

z = −0.67
8.Normal distribution

Answer: x = 259.8 g

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.975 → z = 1.96

Back to x

x = 250 + 1.96·5 = 259.8 g

Answer

x = 259.8 g
9.Confidence intervals

Answer: upper limit = 15.38 h

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Point estimate

x̄₁ − x̄₂ = 107 − 96.5 = 10.5

Standard error

SE = √(8²/30 + 20²/60) = 2.966

Margin of error

E = 1.645·2.966 = 4.88

Interval

10.5 ± 4.88 → [5.62, 15.38]

Answer

upper limit = 15.38 h

The interval does not contain 0, so the means differ significantly.

10.Confidence intervals

Answer: n = 5203

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Solve for n

n = (2.576·28/1)² = 5202

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 5203

Answer

n = 5203
11.Hypothesis tests & confidence intervals

Answer: t = −0.9

Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).

Standard error

SE = 12/√36 = 12/6 = 2

Test statistic

t = (98.2 − 100)/2 = −0.9

Answer

t = −0.9
t0.025; df = 35 = 2.03

|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: z = 1.674

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Sample proportions

p̂₁ = 56/80 = 0.7
p̂₂ = 70/120 = 0.5833
p̂ = 126/200 = 0.63

Standard error under H₀

SE = √(0.63·0.37·(1/80 + 1/120)) = 0.06969

Test statistic

z = (0.7 − 0.5833)/0.06969 = 1.674

Answer

z = 1.674

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

13.Linear regression & correlation

Answer: r = 0.9912

Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.

Means

x̄ = 25/5 = 5
ȳ = 294/5 = 58.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 208
Syy = Σ(y − ȳ)² = 1101

Correlation coefficient

r = 208/√(40·1101) = 0.9912

Answer

r = 0.9912
14.Linear regression & correlation

Answer: e = −4.15

Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.

Means

x̄ = 20/5 = 4
ȳ = 239/5 = 47.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 29

Regression line

b = 29/10 = 2.9
a = 47.8 − 2.9·4 = 36.2
ŷ = 36.2 + 2.9·x

Predicted value

ŷ = 36.2 + 2.9·5.5 = 52.15

Residual

e = 48 − 52.15 = −4.15

Answer

e = −4.15

The point lies below the line.

15.Chi-square tests

Answer: E = 33.81

Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.

Expected count

E = 71·70/147 = 33.81

Answer

E = 33.81
16.Chi-square tests

Answer: χ²crit = 5.991

Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.

Degrees of freedom

df = 3 − 1 = 2

Table

χ²crit(df = 2, α = 0.05) = 5.991

Answer

χ²crit = 5.991

Reject H₀ if the computed χ² is larger than this value.

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